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i have calculated this using "Mathematica" but seems like i a getting negative answer here is what I did Solve[x^5 - 3 x^3 - 3 x == 0, x]. I calculated the value for x and then i integrated the 2 equations from 0 to square root[1/2 (3 + square root[21])] and the answer got was 2 (-(69/8) - (17 square root[21])/8) . The answer is negative and i am sure area is not negative. please help

DeepSea
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shane
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  • See http://www.wolframalpha.com/input/?i=%28%5Cint%5Climits_%7B-sqrt%281%2F2%283%2Bsqrt%2821%29%29%29%7D%5E0+%28%28x%5E5+-+2x%5E3+-+3x%29-+x%5E3+%29+dx%29+%2B+%28%5Cint%5Climits_0%5E%7Bsqrt%281%2F2%283%2Bsqrt%2821%29%29%29%7D+%28x%5E3-%28x%5E5+-+2x%5E3+-+3x%29+%29+dx%29 – AgentS Sep 01 '14 at 01:34
  • draw some pictures to see where $f$ is larger and where $g$ is larger. If both happen, you split the integral into pieces, in effect integrating $|f-g|$ on a bounded interval – Will Jagy Sep 01 '14 at 01:41
  • Try to re-read the question to see if it's readable. Also: see math notation guide. –  Sep 01 '14 at 02:45

1 Answers1

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First you must find the bounds of your integral (where f and g intersect). To do this, set $f(x)=g(x)$. This means $x^{3}=x^{5}-2x^{3}-3x$ or $x(x^{4}-3x^{2}-3)=0.$ Clearly $x=0$ is one place where f and g intersect. Now we can solve $x^{4}-3x^{2}-3=0$ to find the other points where an intersection occurs. Using the quadratic formula we have $$x^{2}=\frac{3\pm\sqrt{21}}{2}$$ We can toss the negative answer because that would imply $x$ is a complex number. This leaves $$x=\pm\sqrt{\frac{3+\sqrt{21}}{2}}$$ and the two roots determine the bounds of your integral, along with $x=0$. For large, negative values of $x$ we know $g(x)<f(x)$ (because g contains a quintic polynomial). This is true until the first point that f and g intersect (i.e. $x=-\sqrt{\frac{3+\sqrt{21}}{2}}$). Now we will have $g(x)>f(x)$ until the next point of intersection at $x=0$. For $x$ where $0 \leq x \leq\sqrt{\frac{3+\sqrt{21}}{2}}$ it swaps back and we have $f(x)>g(x)$. That was a lot of words, but what that means mathematically is: $$A=\int_{-\sqrt{\frac{3+\sqrt{21}}{2}}}^0 \! \left(g(x)-f(x)\right) \, \mathrm{d}x+\int_0^{\sqrt{\frac{3+\sqrt{21}}{2}}} \! \left(f(x)-g(x)\right) \, \mathrm{d}x$$ Even better, we can take advantage of the fact that f and g are odd functions, and that we are only interested in a positive area between the curves to use the integral property $$A=2\int_0^{\sqrt{\frac{3+\sqrt{21}}{2}}} \! \left(f(x)-g(x)\right) \, \mathrm{d}x$$ Now plugging in f and g appropriately yields $$2\int_0^{\sqrt{\frac{3+\sqrt{21}}{2}}} \! \left(-x^{5}+3x^{3}+3x\right) \, \mathrm{d}x$$. I'll leave the final calculation up to you because them integral limits are nasty. The integral itself should be easy to do.

Edit: hopefully you didn't read through all of it yet, I had to fix a couple spots. Should be good now.

graydad
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